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132_ch16 - PHYS 132 WEEK 7 CH 16 16.9 Solve Because the...

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PHYS 132, WEEK 7, CH 16 16.9. Solve: Because the atomic mass number of gold is 197, 1.0 mole of gold has a mass of 197 g or 0.197 kg. The volume of 1.0 mole of gold is V = L 3 = M r Au L = 0.197 kg ( ) 19, 300 kg m 3 1 3 = 2.17 cm 16.12. Solve: Let T F = T C = T : T F = 9 5 T C + 32 ° T = 9 5 T + 32 ° T = - 40 ° That is, the Fahrenheit and the Celsius scales give the same numerical value at - 40 ° . 16.19. Model: Treat the gas in the sealed container as an ideal gas. Solve: (a) From the ideal gas law equation pV = nRT , the volume V of the container is V = nRT p = 2.0 mol ( ) 8.31 J mol K ( ) 273 + 30 ( ) K [ ] 1.013 10 5 Pa = 0.0497 m 3 Note that pressure must be in Pa in the ideal gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is p 1 V T 1 = p 2 V T 2 p 2 = p 1 T 2 T 1 = 1.0 atm ( ) 273 + 130 ( ) K 273 + 30 ( ) K = 1.33 atm Note that gas-law calculations must use T in kelvins. 16.24. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve: (a) The number of moles of oxygen is n = M M mol = 50 g 32 g mol = 1.563 mol (b) The number of molecules is N = nN A = 1.563 mol ( ) 6.02 10 23 mol - 1 ( ) = 9.41 10 23 . (c) The volume of the cylinder V = p r 2 L = p 0.10 m ( ) 2 0.40 m ( ) = 1.257 10 - 2 m 3 .
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